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Odd Behavior of Circuit

Discussion in 'General Electronics Discussion' started by ADavis53, Nov 28, 2017.

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  1. ADavis53

    ADavis53

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    Nov 17, 2017
    Hi all- I have a 12V DC circuit with various components including some relays with built-in timers.

    One particular relay isn't really a relay at all- it's just a timing mechanism. When this relay receives power, it powers the load for 4 seconds and then shuts the load off. The power input voltage to the relay/timer must go away and come back for the load to be powered again.

    I have a 12V DC 85mA alarm bell as the load in this particular case. It works great and as intended.

    I located a much older and larger alarm bell, 12V DC 1A draw and i'm considering using it instead for it's sentimental value... All circuit components can handle the additional current draw, so that is not a consideration.

    The odd thing is that when I put this bell in place of the original bell, the relay seems to power the bell for 8-9 seconds instead of 4. The relay is definitely programmed for 4 seconds. As soon as I swap back to the original bell, the load gets power for the desired 4 seconds again.

    The only obvious difference here is that this second bell draws about 10x the current the first bell does. Not sure why that would impact me here. Could the older bell have stored voltage in it that keeps it ringing after the relay is done delivering power?

    I put my meter on where the (+) and (-) wires for the relay meet the bell and I obviously still have voltage during the full 8-9 seconds the bell is ringing, but I can't obviously say for sure if it's reading voltage off the wires from the relay, the bell or both. If i take the load off and put my meter on the relay wires, i only get voltage for 4 seconds.

    More than anything, i'm curious what could cause this phenomenon.

    Thanks for reading!
    AD
     
  2. 73's de Edd

    73's de Edd

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    Aug 21, 2015
    Certainly sounds as if you're using a common 12 volt power supply for both of units and that with the heavier current consumption of that one bell, that it is pulling down the power supply voltage and upsetting your initial time relays time out constant.
    Check your power supply with the hefty Bell hooked up and see if your voltage is being PULLED down.
    If so and being a wall wart, type of power supply, a heavier current one is being needed.
     
  3. ADavis53

    ADavis53

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    Nov 17, 2017
    Thank you! The power supply is a 12V DC, 3A supply.

    The entire circuit contains the following components that draw current
    * Radio 750mA
    * Relay 1 4mA
    * Relay 2 4mA
    * Hefty Alarm Bell 1A
    * LED Light (200mA) <-- Not currently connected, so factor out for now.

    Even including the light that I have not wired in yet, i'm coming to less than 2A of draw. It's quite possible that the bell draws more than the 1A it says on the label- thing is 40 years old. You seem to be suggesting that this is likely the case and i may need to move up to a 5A (or greater) supply. I'm following.

    Is it common for a 12v/3A supply to fail to provide 12V over short (a foot or two) distances to 2A of load? I calculated my current draw and sized the original power supply based on the math above- maybe I shorted myself on the required current capacity?

    I think what you mean by pulling down, is that if the power supply will provide 12 volts at 3 Amps, that the power supply will provide <12 volts at anything greater than 3 Amps and this is likely what is messing with the relay. Am I interpreting that correctly?

    The relay is designed to operate between 5v and 20v so i wouldn't expect a voltage drop from say 12v to 9v to impact the length of time the timer is active (powering the load). I would however expect it to operate the bell motor more slowly or not shine a 12v light as brightly. Am I off here?

    Thanks again!
     
  4. kellys_eye

    kellys_eye

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    Jun 25, 2010
    The bell may draw an initial high consumption that caused a spike 'low' to occur. Equally such bells are notorious for introducing back EMF (being simple mechanical contacts operating an inductive load) so some form of suppression would be required.
     
  5. Alec_t

    Alec_t

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    Jul 7, 2015
    If the hefty bell is a good olde-fashioned solenoid type then the coil inductance kick-back spike may be causing havoc. Do you have a reverse-biased protection diode across it?

    Edit: Kelly beat me to it!
     
  6. ADavis53

    ADavis53

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    Nov 17, 2017
    Thanks to both- I do not. What should i be looking at to add to my circuit? I'm admittedly not that knowledgeable about circuits.

    Edit: The bell definitely has a large coil of wire acting as a magnet to drive the hammer for the bell- so this is likely the right path.

    I've been doing some reading- would a diode between (+) and (-) after the timer/relay achieve this? See attached image. If so, how do i select the proper diode?

    In the Diagram below the Switch is my relay/timer and the electromagnet is the bell if i'm interpreting properly.
    [​IMG]

    Thanks again!
    AD
     
    Last edited: Nov 28, 2017
  7. 73's de Edd

    73's de Edd

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    1,351
    Aug 21, 2015
    The diode install polarity is being correct and an industry " common as dirt " cheeeeeeeeep priced diode that exceeds the needs by 900% is being a 1N4007.

    Not . . . . . INITIALLY . . . . . even knowing your power supply spec, that sounds adequate.
    I was wondering if voltage reading was at 12 untill the bell rang and then it decreased to 9V or so with the olde tyme heeeeeeveeee-dooodi i i e bell being activated by the circuit.

    On your shown bell circuit, that would only make a single ding upon power application and a fierce continual electromagnet hummmmm until power is removed.
    Assuredley, if yours is making a continual ring, the extra shown portions are being on your unit. The diode would still go directly ACROSS the two coil ends of the electromagnet winding with the diodes bar / cathode being connected to the + battery polarity of the interconnecting wiring.

    upload_2017-11-29_6-42-10.png

    Edd
     
    Last edited: Nov 29, 2017
    Harald Kapp likes this.
  8. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    @73's de Edd : Nice drawing. You sure seem to have a lot of free time, don't you? ;)
    But I dare say you connected the diode backwards: The anode should go to battery - and vice versa:(. The way it is the diode will short circuit the coil.
     
  9. ADavis53

    ADavis53

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    Nov 17, 2017
    Thanks to both! I think i'm following. I'm going to get this thing on it's own power supply / circuit for test purposes and get this figured out- that way i don't let the high voltages potentially created by the magnet destroy my relays or any other components on that circuit not designed for higher voltages. I appreciate the guidance!
     
  10. 73's de Edd

    73's de Edd

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    Aug 21, 2015
    Most Honorable and Esteemed Sir Harald . . . . .Le . . MODERATOR . . .EXTRAORDINAIRE

    but . . .BUT . . .BUT . . . . .BUT . . . . .?

    Nice drawing. (
    Gimme a like )
    Not being all mine . . . just a stock illustration from MIT's Instructors Learning Centers Resources and my adding additional cells to get up to 12VDC and labeling in the damping diode connections as close to the electromagnets coil windings ends, as could be fitted in the drawings vacant areas.
    Really, I also should have moved the "non technical" draftsmans switch placement up to being within the + supply loop at the top.

    You sure seem to have a lot of free time, don't you?
    Sure do, 50 yrs in Mil-Aerospace-Industry then quasi retirement into being a self employed Ind Contractor for lucrative specialized tasks, ( aka . . . .upon payment of my 1K-5K-10K billings, I will then mark my X's on your schematics, after my repairs ) for specialized old military equip or industry. . . . . . .all being jobs opted at my disgression, for the last 20 yrs.

    The anode should go to battery - and vice versa

    Interpreted as . . .diodes negative /anode/triangle as going to circuit negative line.
    " Vice versa " . . . .
    Interpreted as being the diodes positive /cathode/ bar should go to circuit + line.

    With those info bits being relevant and compared to the drawing.
    Strange . . . . . thats exactly the way that I am seeing them on the drawing ?
    Now checking verbiage on the textual content . . . strange . . . .it's also coming out the same?
    The only situation I am seeing is that damper diode is shorting out the high back EMF voltage developed from across the bells electromagnetic coil upon each contact cycles disruption.

    73's de Edd


     
  11. ADavis53

    ADavis53

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    Nov 17, 2017
    I was speaking to someone who knows a little bit about these bells and he was telling me that even back in the day when this bell was made (Guessing late 1970s) that it was common to have a diode for this purpose built in to the bell when it was manufactured.

    Next Question: If the bell is ringing twice as long as i want it to, i can simply cut the timer output time in half if i'm certain that the bell behavior isn't damaging the rest of my circuit. It sounds like a bell like this, if not properly setup could generate high voltages that could damage other components on the circuit not designed for those voltages.

    If, when the bell operates, i am monitoring the voltage between the (+) and the (-) and i never see anything other than 12V, am I safe to assume that this issue was rectified by a built-in diode? Of course, if i see the voltage spike during/after the bell operates, that would be a different story.

    Asked another way: What is the appropriate place to meter for a potentially harmful voltage spike when/after this bell operates?

    Thanks again guys- I'm learning a lot here!
    AD
     
  12. 73's de Edd

    73's de Edd

    3,229
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    Aug 21, 2015
    Why not visually inspect for the presence of a diode being connected across the bells electromagnetic coil proper.
    Without an oscilloscope to be able to observe that coils high voltage back EMF voltage spikes, for only a dollar, would be to go to an ACE . . ."type of " full Service hardware store.
    Find the electrical and repair parts section and get a tiny NE-2 neon lamp with its ~ 2 inch long bare wire leads and a small glass bulb with a tit / tip pulled out on one end and flame sealed.
    If a diode is found across your bells coil, lift one of the diodes leads out of circuit and connect the neon lamp as the diode was connected. Ring the bell and the neon should light up orange for voltage spikes in excess of ~ 60-70 volts. Put your diodes floating lead back in and the voltage spikes should be reduced down to less than a volt and then the neon lamp goes dead.

    73's de Edd
     
  13. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    The way we usually use the symbol is: The cathode is the bar, the anode is the triangle. :rolleyes:

    The schematic in post #7 explicitly says 'diode +' and 'diode -' without reference to the diode symbol (or would I need new glasses?). Which is definitely wrong. 'Diode +' needs to be connected to battery - and vice versa.
     
  14. ADavis53

    ADavis53

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    Nov 17, 2017
    Thanks guys- Of course, the coil itself on this bell is wrapped in some tape... i suppose i will pull it apart and see if i can spot a diode in there.

    Out of fear of destroying the bell- i was trying to avoid tinkering with the bell itself. No other way to detect if there are potentially harmful high voltages by metering via the bell's (+) and (-) wires, correct?
     
  15. BobK

    BobK

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    Jan 5, 2010
    Why not just add the diode and see if it solves the problem? The original use of the bell probably did not incorporate any electronics, and a diode was unnecessary.

    Bob
     
  16. ADavis53

    ADavis53

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    Nov 17, 2017
    Thanks, Bob.

    I suppose for a few pennies there's no reason not to add it.

    here comes the next dumb question: Am i literally inserting the diode between the (+) and (-) wires running to/from the bell such that i end up with the attached image? I presume i'll cut the (+) wire and attach both ends to the cathode end and cut the (-) wire and attach both ends to the anode end- That's it?

    Thanks again to all for bearing with me.
    AD
     

    Attached Files:

  17. Harald Kapp

    Harald Kapp Moderator Moderator

    11,813
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    Nov 17, 2011
    Insert the diode in parallel to the coil, between tge wires where suitable (as close as possible to the coil). You don't have to cut the wire(s). Simply remove some insulation. If it is magnet wire (enamelled wire) it mostly suffices to heat the part where you want to make contact with a hot soldering iron and a blob of molten lead/tin. Once the enamel has burnt away (keep your nose away, too :D) you can attach/solder the diode to the spot.
     
  18. ADavis53

    ADavis53

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    Nov 17, 2017
    Thanks, Harald. I will follow the concept in my previous post but slide the diode all the way up as close to the bell as i feasibly can. I will let you know how this impacts the behvaior of the bell.
     
  19. ADavis53

    ADavis53

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    Nov 17, 2017
    No dice, guys. I picked up some 1N4007 diodes and installed per diagram above and I still get the same result- the bell is resetting the timer on my relay and ringing continuously.

    I'm attaching a couple pics to show how I put the diode on. Hoping I made a simple mistake and you guys can straighten me out.

    Best
    Alex
     

    Attached Files:

  20. mr fixit

    mr fixit

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    Jun 23, 2013
    Hi Alex.

    IF the twin lead goes to the bell coil/s and the red lead is the positive, black lead is negative, to your power supply, then yes. That diode is wired correctly, although it would be more effective if it was connected as close as possible to the coil.
    Have you measured the +12 Volts when the bell is ringing? Is it 12Volts?
    You could try this:
    Measure the power supply voltage with nothing connected to it, just switched on.
    Then connect ONLY the bell to the power supply. Yes, it will be ringing.
    Measure the volts now... Still 12 Volts?
    If its above 12 volts (well above) with nothing connected to it, then way less than 12 volts when the bell is connected, I would suggest that's the problem.
    To prove that theory once and for all, you could try a different power supply. A car battery comes to mind!

    If that's not the problem then there is so much hash coming from that old bell that its upsetting the timing circuit.
    Then it will be time to get serious. You will need to "decouple" the power supply that supplies the timing circuit from that which supplies the bell. And/or employ more effective filtering to the bell coil than simply a reverse diode.
    What is the timing circuit? Do you have a diagram?
     
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